状压 DP：[USACO06NOV] Corn Fields，[USACO13NOV] No Change

[USACO06NOV] Corn Fields

（试题来源：Link ）

题目描述

Farmer John has purchased a lush new rectangular pasture composed of $M$ by $N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12)$ square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

农场主 John 新买了一块长方形的新牧场，这块牧场被划分成 $M$ 行 $N$ 列 $(1 ≤ M ≤ 12; 1 ≤ N ≤ 12)$，每一格都是一块正方形的土地。John 打算在牧场上的某几格里种上美味的草，供他的奶牛们享用。

遗憾的是，有些土地相当贫瘠，不能用来种草。并且，奶牛们喜欢独占一块草地的感觉，于是 John 不会选择两块相邻的土地，也就是说，没有哪两块草地有公共边。

John 想知道，如果不考虑草地的总块数，那么，一共有多少种种植方案可供他选择？（当然，把新牧场完全荒废也是一种方案）

输入输出格式

输入格式：

第 $1$ 行：两个整数 $M$ 和 $N$，用空格隔开。
第 $2$ 到第 $M+1$ 行：每行包含 $N$ 个用空格隔开的整数，描述了每块土地的状态。第 $i+1$ 行描述了第 $i$ 行的土地，所有整数均为 $0$ 或 $1$，是 $1$ 的话，表示这块土地足够肥沃，$0$ 则表示这块土地不适合种草。

输出格式：

一个整数，即牧场分配总方案数除以 $100,000,000$ 的余数。

输入输出样例

输入样例 #1：

输出样例 #1：

状态压缩 DP，预处理每行可能的全部状态。每个当前状态由上一行所有并行状态累加得到。

/* [USACO06NOV] Corn Fields

 * Au: GG

 */

#include <cstdio>

const int N = 4197, M = 15, MOD = 100000000;

int n, m, dp[M][N], ans;

struct node {

    int s[N], st;

} p[M];

int main() {

    scanf("%d%d", &m, &n);

    for (int i=1, a, t; i<=m; i++) {

        t = 0;

        for (int j=1; j<=n; j++)

            scanf("%d", &a), t = (t<<1) + 1 - a;

        for (int j=0; j< (1<<n); j++)

            if ((j&(j<<1))||(j&(j>>1))||(j&t)) continue;

            else p[i].s[++p[i].st] = j;

    }

    for (int i=1; i<=p[1].st; i++) dp[1][i] = 1;

    for (int i=2; i<=m; i++)

        for (int j=1; j<=p[i].st; j++)

            for (int k=1; k<=p[i-1].st; k++)

                if (!(p[i].s[j] & p[i-1].s[k])) dp[i][j] += dp[i-1][k];

    for (int i=1; i<=p[m].st; i++)

        ans = (ans + dp[m][i]) % MOD;

    printf("%d\n", ans);

    return 0;

}

[USACO13NOV] No Change

（试题来源：Link ）

题目描述

Farmer John is at the market to purchase supplies for his farm. He has in his pocket $K$ coins $(1 \leq K \leq 16)$, each with value in the range $1..100,000,000$. FJ would like to make a sequence of $N$ purchases $(1 \leq N \leq 100,000)$, where the ith purchase costs $c(i)$ units of money $(1 \leq c(i) \leq 10,000)$. As he makes this sequence of purchases, he can periodically stop and pay, with a single coin, for all the purchases made since his last payment (of course, the single coin he uses must be large enough to pay for all of these). Unfortunately, the vendors at the market are completely out of change, so whenever FJ uses a coin that is larger than the amount of money he owes, he sadly receives no changes in return!

Please compute the maximum amount of money FJ can end up with after making his $N$ purchases in sequence. Output $-1$ if it is impossible for FJ to make all of his purchases.

John 到商场购物，他的钱包里有 $K(1 \leq K \leq 16)$ 个硬币，面值的范围是 $1..100,000,000$。

John 想按顺序买 $N$ 个物品 $(1 \leq N \leq 100,000)$，第 $i$ 个物品需要花费 $c(i)$ 块钱，$(1 \leq c(i) \leq 10,000)$。

在依次进行的购买 $N$ 个物品的过程中，John 可以随时停下来付款，每次付款只用一个硬币，支付购买的内容是从上一次支付后开始到现在的这些所有物品（前提是该硬币足以支付这些物品的费用）。不幸的是，商场的收银机坏了，如果 John 支付的硬币面值大于所需的费用，他不会得到任何找零。

请计算出在购买完 $N$ 个物品后，John 最多剩下多少钱。如果无法完成购买，输出 $- 1$

输入输出格式

输入格式：

Line $1$: Two integers, $K$ and $N$.
Lines $2..1+K$: Each line contains the amount of money of one of FJ's coins.
Lines $2+K..1+N+K$: These $N$ lines contain the costs of FJ's intended purchases.

输出格式：

Line $1$: The maximum amount of money FJ can end up with, or $-1$ if FJ cannot complete all of his purchases.

输入输出样例

输入样例 #1：

输出样例 #1：

说明

FJ has 3 coins of values 12, 15, and 10. He must make purchases in sequence of value 6, 3, 3, 2, 3, and 7.

FJ spends his 10-unit coin on the first two purchases, then the 15-unit coin on the remaining purchases. This leaves him with the 12-unit coin.

前缀和存物品费用；$dp[i]$ 为 $i$ 状态能购买的最大数量商品。由每个状态产生新的状态（刷表法），二分求出当前已使用 $i$ 状态硬币之后，下一个硬币能购买到的最大数量商品，统计最大值。

二分理论上可以用 lower_bound 替换，可惜本人技艺不精，bug 太多故未使用。

/* [USACO13NOV] No Change

 * Au: GG

 */

#include <cstdio>

#include <algorithm>

using namespace std;

const int N = 100005;

int k, n, w[N], coin[18], dp[1<<17], ans=-1;

int main() {

    scanf("%d%d", &k, &n);

    for (int i=1; i<=k; i++) scanf("%d", &coin[i]);

    for (int i=1; i<=n; i++) scanf("%d", &w[i]), w[i]+=w[i-1];

    for (int i=0; i<(1<<k); i++) {

        if (dp[i]==n) {

            int res=0;

            for (int j=1; j<=k; j++) if (!(i&(1<<j-1))) res+=coin[j];

            ans = max(ans, res);

        }

        for (int j=1; j<=k; ++j) if (!(i&(1<<j-1))) {

            int now = i | (1<<j-1);

            int l=dp[i], r=n, mid;

            while (l<r) {

                mid = (l+r+1)>>1;

                if (w[mid]-w[dp[i]]<=coin[j]) l=mid; else r=mid-1;

            }

            dp[now] = max(dp[now], l);

        }

    }

    printf("%d\n", ans);

    return 0;

}